Let $g$ be a vector-valued function defined by $g(t)=(\sin(2 t+1),5\cos(4-t))$. Find $g$ 's second derivative $g''(t)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\left(-2\cos(2t+1),5\sin(4-t)\right)$ (Choice B) B $(-4\sin(2t+1),-5\cos(4-t))$ (Choice C) C $(2\sin(2t+1),-5\cos(4-t))$ (Choice D) D $-2\cos(2t+1)+5\sin(4-t)$
Solution: We are asked to find the second derivative of $g$. This means we need to differentiate $g$ twice. In other words, we differentiate $g$ once to find $g'$, and then differentiate $g'$ (which is a vector-valued function as well) to find $g''$. Recall that $g(t)=\left(\sin(2 t+1),5\cos(4-t)\right)$. Therefore, $g'(t)=\left(2\cos(2t+1),5\sin(4-t)\right)$. Now let's differentiate $g'(t)=(2\cos(2t+1),5\sin(4-t))$ to find $g''$. $g''(t)=(-4\sin(2t+1),-5\cos(4-t))$ In conclusion, $g''(t)=\left(-4\sin(2t+1),-5\cos(4-t)\right)$.